Question

If $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$ and $B = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}$ are arbitrary vectors in $\mathbb{R}^{2 \times 2}$, then the mapping $(A, B) = a_{11}b_{11} + a_{12}b_{12} + a_{21}b_{21} + a_{22}b_{22}$ defines an inner product in $\mathbb{R}^{2 \times 2}$. Use this inner product to determine $(A, B)$, $||A||$, $||B||$, and the angle $\alpha_{A, B}$ between $A$ and $B$ for $A = \begin{bmatrix} -4 & -4 \\ 4 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -5 & -1 \\ -3 & -5 \end{bmatrix}$. $(A, B) = ||A|| = ||B|| = $\alpha_{A, B} =

          If
$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$ and $B = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}$
are arbitrary vectors in $\mathbb{R}^{2 \times 2}$, then the mapping
$(A, B) = a_{11}b_{11} + a_{12}b_{12} + a_{21}b_{21} + a_{22}b_{22}$
defines an inner product in $\mathbb{R}^{2 \times 2}$. Use this inner product to determine $(A, B)$, $||A||$, $||B||$, and the angle $\alpha_{A, B}$
between $A$ and $B$ for
$A = \begin{bmatrix} -4 & -4 \\ 4 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -5 & -1 \\ -3 & -5 \end{bmatrix}$.
$(A, B) = 
||A|| = 
||B|| = 
$\alpha_{A, B} =
        
Show more…
If
A = 
    < b m a t r i x > and B = 
    < b m a t r i x >
are arbitrary vectors in ℝ^2 × 2, then the mapping
(A, B) = a11b11 + a12b12 + a21b21 + a22b22
defines an inner product in ℝ^2 × 2. Use this inner product to determine (A, B), ||A||, ||B||, and the angle αA, B
between A and B for
A = 
    < b m a t r i x > and B = 
    < b m a t r i x >.
(A, B) = 
||A|| = 
||B|| =αA, B =

Added by Arthur R.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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If A = [a11 a12] and B = [a21 a22] are arbitrary vectors in IR^2, then the mapping (A, B) = a11b11 + a12b12 + a21b21 + a22b22 defines an inner product in IR^2. Use this inner product to determine (A, B), |A|, |B|, and the angle θ between A and B for A, B ∈ IR^2.
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Transcript

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00:01 Hello everyone we are going to solve a question in this question we are given a matrix which is equals to 5 minus 2 4 minus 2 and matrix b which is equal to minus 5 minus 2 1 and 1 and in a part in first part of this question we have to find a comma b now as we know this is equal to a11 multiplied with b11 plus a12 multiplied with b12 plus a211 plus a21 multiplied with b12 plus a2222 by putting the value this is equal to 5 multiplied with minus 5 plus minus 2 multiplied with minus 2 plus 4 multiplied with 1 plus minus 2 multiplied with 1.
00:58 Now on solving this is equals to minus 25 plus 4 plus 4 and minus 2 which is equals to minus 19.
01:08 So the answer for a part that is this is equals to minus 19 which is the final answer for first part.
01:17 Moving on to the second part in this part we have to find the norm of a.
01:24 Now as we have formula that this is equals to.
01:28 Under root of a comma a.
01:32 On solving this we have this is equals to under root minus 5 square sorry this is plus 5 square plus minus 2 square plus 4 square plus minus 2 square on solving this is equals to under root 25 plus 4 plus 16 plus 4 so this is equals to under root 25 plus 16 plus 4 so this is which is equal to 7.
02:00 So hence we have this norm a is equals to 7 which is the final answer for second part.
02:07 Moving on to the third one in this we have to find norm b...
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